Question

Find the asymptotes of the following hyperbolas and also the equations to their conjugate hyperbolas.
$55x^2-120xy+20y^2+64x-48y=0.$

Answer 1

Given equation of the hyperbola: $55x^2-120xy+20y^2+64x-48y=0$

The equation of a hyperbola and the combined equation of the asymptotes differ only in the constant term.

Therefore the combined equation of the asymptotes is of the form: $55x^2-120xy+20y^2+64x-48y+K=\mathrm{0...}\left(i\right)$

Factorising the 2nd degree terms, gives

$55x^2-120xy+20y^2=55x^2-110xy-10xy+20y^2=(x-2y)(55x-10y)$

Therefore the asymptotes have equations $x-2y+l=0$ and $55x-10y+m=0$ and their combined equation will be $(x-2y+l)(55x-10y+m)=\mathrm{0...}\left(ii\right)$.

Equations (i) & (ii) represent the same pair of lines.

Comparing the coefficient of x, y and constant term gives,

$55l+m=64,-2m-10l=-48$ & $lm=K$,

which gives $l=\frac{4}{5}$, and $m=20$

Therefore the asymptotes are $x-2y+\frac{4}{5}=0$ and $55x-10y+20=0$ and their combined equation is $55x^2-120xy+20y^2+64x-48y+16=0$.

Now using the fact that: equation of hyperbola + equation of conjugate hyperbola = 2x(equation of asymptotes)

$\Rightarrow$ equation of conjugate hyperbola = 2x(equation of asymptotes) - equation of hyperbola

$\Rightarrow$ equation of conjugate hyperbola = $2(55x^2-120xy+20y^2+64x-48y+16)-(55x^2-120xy+20y^2+64x-48y)$

$\Rightarrow$ equation of conjugate hyperbola = $55x^2-120xy+20y^2+64x-48y+32=0$